Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{5y^2 - 9y}{6y} \div \dfrac{3(5y - 9)}{-6} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{5y^2 - 9y}{6y} \times \dfrac{-6}{3(5y - 9)} $ When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ (5y^2 - 9y) \times -6 } { 6y \times 3(5y - 9) } $ $ z = \dfrac {-6 \times y(5y - 9)} {6y \times 3(5y - 9)} $ $ z = \dfrac{-6y(5y - 9)}{18y(5y - 9)} $ We can cancel the $5y - 9$ so long as $5y - 9 \neq 0$ Therefore $y \neq \dfrac{9}{5}$ $z = \dfrac{-6y \cancel{(5y - 9})}{18y \cancel{(5y - 9)}} = -\dfrac{6y}{18y} = -\dfrac{1}{3} $